The answer should be D all of the above
To solve for the empirical formula, we write first all the data.
Given:
Compound 1: 76 wt% Ru and 24wt% O
Compound 2: 61.2 wt% Ru and 38.8 wt% O
Required: Empirical Formula of Compound 1
Solution:
Assume total mass of the compound is 100 g
Solving for Compound 1,
76 g Ru x <u>1 mol Ru </u> = 0.75195 mol Ru
101.07 g Ru
24 g O x <u>1 mol O </u> = 1.5 mol O
16 g O
Then, divide each mole with the smallest number of moles calculated
Ru = 0.75195 mol/0.75195 mol = 1
O = 1.5 mol/0.75195 mol = 2
Therefore, the empirical formula for Compound 1 is RuO2.
<em>ANSWER: RuO2</em>
Correct answer: C) 1.6 g
Molarity is the moles of solute present per liter solution.
The given molarity of the solution is 0.01 M.
Volume of the solution = 1 L
Calculating the moles from molarity and volume:
Converting moles to mass of copper sulfate:
